Mechanics is fun! (3)

The particle's velocity before the collision is $(0,-v)$ , and its velocity after the collision is $(-\tfrac12v,0)$ . Thus the particle experiences an impulse $\mathbf{I} = \tfrac12mv(-1,2)$ . A unit vector normal to the slope of the wedge is $\mathbf{n} = (-\tfrac35,\tfrac45)$ , while a unit vector parallel to the slope is $\mathbf{t} = (\tfrac45,\tfrac35)$ , and we see that $\mathbf{I} \cdot \mathbf{n} = \tfrac{11}{10}mv$ and $\mathbf{I}\cdot\mathbf{t} = \tfrac15mv$ . Thus the particle experiences an impulse of $\tfrac{11}{10}mv$ normal to the slope, and one of $\tfrac15mv$ parallel to the slope. Thus the coefficient of friction between the particle and the slope is at least $\tfrac{2}{11}$ .

If the slope's velocity after the collision is $(V,0)$ , then the speed of approach of the particle and the wedge is $\mathbf{n}\cdot(0,v) \, = \, \tfrac45v$ , while the speed of separation of the particle and the wedge after the collision is $-\mathbf{n}\cdot(V+\tfrac12v,0) \,=\, \tfrac35(V + \tfrac12v)$ . Since the coefficient of restitution is $\tfrac{9}{16}$ , we have $\tfrac35(V + \tfrac12v) \; = \; \tfrac{9}{16} \times \tfrac45v$ and hence $V = \tfrac14v$ . In addition to an impulse of $-\mathbf{I}$ from the particle, the wedge receives an impulse of $\mathbf{J}$ from the ground, and $-\mathbf{I} + \mathbf{J} \; =\; \tfrac14mv(1,0)$ so that $\mathbf{J} \; = \; mv(-\tfrac14,1)$ Thus the wedge experiences a normal impulse of $mv$ , and a frictional impulse of $\tfrac14mv$ , from the ground, and hence (since friction is limiting in this case) the coefficient of friction between the wedge and the ground is $\tfrac14$ .

Thus $A \,=\, \tfrac{2}{11} + \tfrac14 \,=\, \tfrac{19}{44}$ , and $100A = 43.1818181818\ldots$ . The answer is $\boxed{43}$ .